Answer
$\dfrac{\pi}{24} (1-\cos 1)$
Work Step by Step
Consider $u=3x$ and $v=2y$
Here, we have: $Jacobian (x,y)=\begin{vmatrix} \dfrac{\partial x}{\partial u}&\dfrac{\partial x}{\partial v}\\\dfrac{\partial y}{\partial u}&\dfrac{\partial y}{\partial v}\end{vmatrix}=\begin{vmatrix} \dfrac{1}{3}&0\\0&\dfrac{1}{2}\end{vmatrix}=\dfrac{1}{6}$
Now, $\iint_R \sin(9x^2+4y^2) dx dy=(\dfrac{1}{6}) \iint_{R} \sin (u^2+v^2) |J| du dv$
or, $(\dfrac{1}{6}) \int_0^{\pi/2} \int_0^1 \sin (r^2) r dr d
theta =(\dfrac{1}{6}) [\theta]_0^{\pi/2}[(-1/2) \cos (r^2)]_0^1$
Thus, we have
$\iint_R \sin(9x^2+4y^2) dx dy= \dfrac{-1}{12}(\pi/2) (\cos (1)-1)=\dfrac{\pi}{24} (1-\cos 1)$
